Physics, 18.03.2021 02:00 catherine51

What is the electric resistance of a copper wire 0.65 m long with a diameter of 2 mm?

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Answer from: binodkharal2048

To calculate resistance, we will use the formula:

R = ρl / A

where ρ(rho) is the resistivity of the material, l is the length of the wire and A is the cross-sectional area

We are given:

Length of wire (l) = 0.65 m

Diameter of wire (d) = 2 mm

Material of wire: Copper

Some important conversions:

- radius of the wire = diameter/2

radius = 2 mm/2 = 1 mm OR 1 * 10⁻³m

Calculating the cross-sectional area:

Cross-sectional area is the area of the circle at the end of the wire.

Cross-sectional area = π(r²)

Area = π(1 * 10⁻³)² [replacing the value of r]

Area = π * 10⁻⁶ m²

Calculating the resistance:

using the formula mentioned before:

R = ρl / A

R = $(1.7 * 10^{-8}ohm*m ) * \frac{0.65 m}{3.14 * 10^{-6}m^{2}}$ [resistivity of copper = 1.7 × 10⁻⁸ Ωm]

R = 3.52 * 10⁻³ (approx)

Answer from: zanaplen27

3.5 × 10⁻³ Ω

Explanation:

The resistance of a conductor is calculated by the formula:

$\displaystyle R=\rho \cdot \frac{l}{A}$ where R = electric resistance (Ω), ρ = resistivity (Ω · m), l = length of the wire (m), A = area of the cross-section of the wire (m²)

Let's start by converting the diameter of the copper wire to meters.

2 mm → .002 m

Since we want the radius of the cross-section, we will divide .002 m by 2.

.002/2 = .001 m

The radius of the copper wire is .001 m. We can calculate the area of the circular cross-section by using the formula:

$A=\pi r^2$ $A= \pi (.001)^2$ $A= \pi \cdot 10^-^6$

The area of the cross-section is π · 10⁻⁶ m².

The length of the wire is 0.65 m long. We do not have to convert units for the length of the wire since it is already in the SI units: meters.

Assuming the copper wire is at 20°C, we know that its resistivity is 1.7 · 10⁻⁸ Ω · m.

Using these three variables, we can solve for R in the formula for electric resistance.

ρ = 1.7 · 10⁻⁸ Ω · m l = 0.65 m A = π · 10⁻⁶ m²

Substitute these values into the equation.

$\displaystyle R=\rho \cdot \frac{l}{A}$ $\displaystyle R=(1.7 \cdot 10^-^8\ \Omega \cdot \text{m}) \cdot \frac{0.65 \ \text{m}}{(\pi \cdot 10^-^6 \ \text{m}^2 )}$ $\displaystyle R=\frac{(1.105 \cdot 10^-^8 \ \Omega \cdot \text{m}^2)}{(\pi \cdot 10^-^6 \ \text{m}^2)}$ $R=.3517324242 \cdot 10^-^2 \ \Omega$ $R=.003517324242 \ \Omega$

Notice how the unit m² cancels out, leaving us with Ω (units of electrical resistance).

The electric resistance of a copper wire 0.65 m long with a radius of .001 m is .0035 ohms (Ω), or 3.5 × 10⁻³ Ω.

Answer from: Quest

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.

Answer from: Quest

Iam looking for this answer too!

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